Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

Recall Newton's Law of Gravitation, which asserts that the magnitude \(F\) of the force of attraction between objects of masses \(M\) and \(m\) is \(F=G M m / r^{2}\), where \(r\) is the distance between them and \(G\) is a universal constant. Let an object of mass \(M\) be located at the origin, and suppose that a second object of changing mass \(m\) (say from fuel consumption) is moving away from the origin so that its position vector is \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\). Obtain a formula for \(d F / d t\) in terms of the time derivatives of \(m\), \(x, y\), and \(z\).

Step-by-Step Solution

Answer

The formula for \( \frac{dF}{dt} \) is \[ G M \left( \frac{dm}{dt} \cdot \frac{1}{r^2} - 2m \cdot \frac{x\dot{x} + y\dot{y} + z\dot{z}}{r^4} \right) \].

Step 1: Write the original formula for the gravitational force

The gravitational force between two objects is given by Newton's Law of Gravitation: \[ F = \frac{G M m}{r^2} \] where \( F \) is the force, \( G \) is the gravitational constant, \( M \) and \( m \) are the masses, and \( r \) is the distance between them.

Step 2: Express distance in terms of the position vector

The distance \( r \) between the two objects is the magnitude of the position vector \( \mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \). Therefore, \[ r = \sqrt{x^2 + y^2 + z^2} \] This gives us the distance in terms of the coordinates \( x, y, \) and \( z \).

Step 3: Differentiate the force with respect to time

We need to find the derivative \( \frac{dF}{dt} \). Using the given formula for \( F \), we differentiate:\[ \frac{dF}{dt} = \frac{d}{dt} \left( \frac{G M m}{r^2} \right) \]. This requires using the product and chain rule of differentiation.

Step 4: Apply the product rule to differentiate

Applying the product rule, \[ \frac{dF}{dt} = G M \left( \frac{dm}{dt} \cdot \frac{1}{r^2} + m \cdot \frac{d}{dt}\left( \frac{1}{r^2} \right) \right) \]. This expression accounts for the change in mass \( m \) and the change in position \( \mathbf{r} \).

Step 5: Differentiate \( \frac{1}{r^2} \) using the chain rule

To find \( \frac{d}{dt} \left( \frac{1}{r^2} \right) \), apply the chain rule:\[ \frac{d}{dt} \left( \frac{1}{r^2} \right) = -\frac{2}{r^3} \frac{dr}{dt} \], since \( \frac{d}{dt} \left( r^{-2} \right) \) involves the derivative of \( r \) with respect to time.

Step 6: Calculate the time derivative of \( r \)

The time derivative \( \frac{dr}{dt} \) is defined by:\[ \frac{dr}{dt} = \frac{1}{2\sqrt{x^2 + y^2 + z^2}} \cdot 2 \left( x\frac{dx}{dt} + y\frac{dy}{dt} + z\frac{dz}{dt} \right) = \frac{x\dot{x} + y\dot{y} + z\dot{z}}{r} \]. Here, \( \dot{x}, \dot{y}, \dot{z} \) denote the derivatives of \( x, y, z \) with respect to time.

Step 7: Combine results to find \( \frac{dF}{dt} \)

Putting it all together, we substitute \( \frac{d}{dt} \left( \frac{1}{r^2} \right) \) and \( \frac{dr}{dt} \): \[ \frac{dF}{dt} = G M \left( \frac{dm}{dt} \cdot \frac{1}{r^2} - 2m \cdot \frac{1}{r^3} \cdot \frac{x\dot{x} + y\dot{y} + z\dot{z}}{r} \right) \]. This is the expression for the rate of change of the gravitational force.

Key Concepts

Newton's Law of Gravitation

Newton's Law of Gravitation is a fundamental principle formulated by Isaac Newton, describing the attractive force between two masses. It states that every particle in the universe attracts every other particle with a force that is proportional to the product of their masses. The force is inversely proportional to the square of the distance between their centers. In mathematical terms, the gravitational force \( F \) can be expressed as:
\[ F = \frac{G M m}{r^2} \]
where:

• \( F \) is the force of attraction between the two masses.
• \( G \) is the gravitational constant, a universal constant with a value of approximately \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \).
• \( M \) and \( m \) are the masses of the objects.
• \( r \) is the distance between the centers of the two masses.

Understanding this law provides insight into how massive objects like planets and stars exert gravitational pull on each other, influencing the motion of celestial bodies across the cosmos.

Time Derivatives

In calculus, a derivative represents how a function changes as its input changes; a time derivative expresses this change with respect to time. It gives the rate at which a quantity, such as position, changes over time, often denoted by a dot over the variable. In physics problems involving forces and motion, time derivatives play a crucial role. They help us describe how physical quantities evolve, and they are essential for understanding dynamics.
For instance, the time derivative of a position vector \( \mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \) is the velocity vector of the moving object:
\( \mathbf{v} = \dot{x} \mathbf{i} + \dot{y} \mathbf{j} + \dot{z} \mathbf{k} \)
This expression shows that each component of the position changes over time, represented by \( \dot{x}, \dot{y}, \text{and } \dot{z} \). In the context of forces like gravity, time derivatives give crucial insight into how altering position, velocity, or mass influences the forces acting on objects.

Product Rule of Differentiation

The product rule is a fundamental tool in calculus used to take the derivative of a product of two functions. When functions are dependent on variables like time, this rule helps in finding how such relationships evolve over time. The product rule states that if you have two functions \( u(t) \) and \( v(t) \), their product \( u(t) \cdot v(t) \) differentiates as:
\[ \frac{d}{dt}[u(t) \cdot v(t)] = \frac{du}{dt} \cdot v(t) + u(t) \cdot \frac{dv}{dt} \]
This rule allows us to differentiate complex relationships where multiple quantities change simultaneously. For example, while calculating \( \frac{dF}{dt} \) for gravitational forces, where both mass and distance change over time, the product rule is indispensable. It effectively captures these concurrent changes:

• Identify the product of the two changing factors (e.g., \( m \) and \( 1/r^2 \)).
• Differentiate each component based on how they change over time.
• Combine these derivatives to determine the overall rate of change.

By using the product rule, we gain a comprehensive view of how multiplying factors contribute to the rate of change of a physical quantity, ensuring accurate results in calculations involving dynamic systems.